Question

(a) What is the energy stored in the 10.0 μF capacitor of a heart defibrillator charged to 9.00×103V ? (b) Find the amount of stored charge.

Answer 1

Answer:

(a) $U=405J$

(b) $Q=0.09C$

Explanation:

(a) The potential energy stored in a capacitor is given by the expression:

$U=\frac{QV}{2}$

Q is the stored charge and V the potential difference between capacitor plates. In a capacitor, we have:

$Q=CV$

Replacing this in the energy equation:

$U=\frac{CV^2}{2}\\U=\frac{10*10^{-6}F(9*10^3V)^2}{2}\\U=405J$

(b) Using the energy expression and solving for Q:

$U=\frac{QV}{2}\\Q=\frac{2U}{V}\\Q=\frac{2(405J)}{9*10^3V}\\\\Q=0.09C$